KindiakMath

Equating Ranges and Kernels

Let V be a vector space and T : V \to V be a linear transformation.

Problem 1. Suppose T^2 = I, where we denote I = I_V for brevity. Prove that (T-I)(V) = \mathrm{ker}(T+I).

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Solution. Under the hypothesis that T^2 = I, we have (T+I)(T-I) = O.

We will first prove (\subseteq). Fix v \in (T-I)(V). Find w \in V such that v = (T-I)(w). Then

(T+I)(v) = (T+I)(T-I)(w) = (T^2 - I)(w) = O(w) = 0,

so that v \in \mathrm{ker}(T+I). Hence, (T-I)(V) \subseteq \mathrm{ker}(T+I).

Now, we prove (\supseteq). Fix v \in \mathrm{ker}(T+I). Then

0 = (T+I)(v) = T(v) + v \quad \Rightarrow \quad v = -T(v) = T(-v).

By repeated application of this result,

v = \frac 12v + \frac 12v = T(-\frac 12 v) - (-\frac 12 v) = (T-I)(-\frac 12 v) \in (T-I)(V).

Hence, \mathrm{ker}(T+I) \subseteq (T-I)(V) .

Combining both results, (T-I)(V) = \mathrm{ker}(T+I).

Problem 2. Suppose T^2 = T. Prove that

V = \ker(T) + T(V),\quad \ker(T) \cap T(V) = \{ 0\}.

In this case, we call T a projection operator. In particular, if T is the orthogonal projection, we obtain orthogonal decomposition, where all vectors in \ker(T) are “perpendicular” or orthogonal to vectors in T(V).

(Click for Solution)

Solution. Fix v \in V. Write

v = (v - T(v)) + T(v).

Since

\begin{aligned} T(v - T(v)) & = T(v) - T^2(v)\\ &= T(v) - T(v) \\& = 0, \end{aligned}

we have v - T(v) \in \ker(T), so that

v = \underbrace{ (v - T(v)) }_{\in\ker(T)} + \underbrace{ T(v) }_{\in T(V)} \in \ker(T) + T(V).

To establish the trivial intersection, fix v \in \ker(T) \cap T(V). Since v \in T(V), find u \in V such that v = T(u):

v = T(u) = T^2(u) = T(v).

Since v \in \ker(T), v = T(v) = 0. Therefore, \{0\} \subseteq \ker(T) \cap T(V) \subseteq \{0\}.

—Joel Kindiak, 29 Nov 24, 2130H

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